17—Densities and Distributions 424The expression−∂(1/r)/∂zis such a simple way to compute this, the potential of an electric
dipole, that it is worth trying to understand why it works. And in the process,whyis this an electric
dipole? The charge density, the source of the potential is a derivative, and a derivative is (the limit of)
a difference quotient. This density is just that of two charges, and they produce potentials just as in
Eq. (17.44). There are two point charges, with delta-function densities, one at the origin the other at
−zˆ∆z.
ρ=
−p
∆z
[
δ(~r+ˆz∆z)−δ(~r)
]
gives potential−p
4 π 0 ∆z
[
1
|~r+zˆ∆z|
−
1
r
]
(17.48)
P
r 1 =r
r 2
q
−q
q r^1
r 2
P 2
P 1
(a) (b)The picture of the potential that arises from this pair of charges is (a). A negative charge (−q=
−p/∆z) at−zˆ∆zand a corresponding positive charge+qat the origin. This picture explains why this
charge density,ρ(~r) =−p∂δ(~r)/∂z, represent an electric dipole. Specifically it represents a dipole
whose vector representation points toward+zˆ, from the negative charge to the positive one. It even
explains why the result (17.47) has the sign that it does: The potentialφis positive along the positive
z-axis and negative below. That’s because a point on the positivez-axis is closer to the positive charge
than it is to the negative charge.
Now the problem is to understandwhythe potential of this dipole ends up with a result as simple
as the derivative(−p/ 4 π 0 )∂(1/r)/∂z: The potential at the point P in the figure (a) comes from the
two charges at the two distancesr 1 andr 2. Construct figure (b) by moving the liner 2 upward so that
it starts at the origin. Imagine a new charge configuration consisting of onlyonecharge,q=p/∆zat
the origin. Now evaluate the potentials that this single charge produces at two different points P 1 and
P 2 that are a distance∆zapart. Then subtract them.
φ(P 2 )−φ(P 1 ) =
q
4 π 0
[
1
r 2
−
1
r 1
]
In the notation of the preceding equations this is
φ(P 2 )−φ(P 1 ) =
q
4 π 0
[
1
|~r+zˆ∆z|
−
1
r
]
=
p
4 π 0 ∆z
[
1
|~r+zˆ∆z|
−
1
r
]
Except for a factor of (− 1 ), this is Eq. (17.48), and it explains why the potential caused by an ideal
electric dipole at the origin can be found by taking the potential of a point charge and differentiating
it.
φpoint charge=
q
4 π 0 r
φdipole=−a
∂
∂z
φpoint charge
Here, the electric dipole strength isqa.
Can you repeat this process? Yes. Instead of two opposite point charges near each other, you
can place two opposite point dipoles near each other.